Continuous Functional Equation

Detailed Solution

Since $f:\mathbb R\to\mathbb R$ is given to be continuous, its restriction to every compact interval $[a,b]$ is also continuous. By the Extreme Value Theorem, every continuous real-valued function on a compact interval is bounded. Therefore, $f$ must be bounded on every compact interval. Hence, statement D is necessarily true. Now check the other options. For A, the functional equation does not force $f$ to be linear. For example, define $$ f(x)= \begin{cases} x\left(1+\dfrac12\sin\left(2\pi\log_2|x|\right)\right),&x\ne0,\\[6pt] 0,&x=0. \end{cases} $$ This function is continuous at every nonzero point. Also, because the factor multiplying $x$ is bounded, $f(x)\to0$ as $x\to0$, so it is continuous at $0$. Moreover, $$ \begin{aligned} f(2x) &=2x\left(1+\dfrac12\sin\left(2\pi\log_2|2x|\right)\right)\\ &=2x\left(1+\dfrac12\sin\left(2\pi(\log_2|x|+1)\right)\right)\\ &=2x\left(1+\dfrac12\sin\left(2\pi\log_2|x|\right)\right)\\ &=2f(x), \end{aligned} $$ because $\sin(\theta+2\pi)=\sin\theta$. Thus the functional equation can have nonlinear continuous solutions, so A need not be true. For B, consider $$ f(x)=|x|. $$ It is continuous and satisfies $f(0)=0$. Also, $$ f(2x)=|2x|=2|x|=2f(x). $$ However, $$ f(-x)=|-x|=|x|=f(x), $$ so $f$ is even, not odd. Hence B need not be true. For C, the equation relates values of the function along dyadic scalings: $$ f(2x)=2f(x), $$ and more generally, $$ f(2^n x)=2^n f(x) $$ for every integer $n$. But continuity imposes conditions throughout $\mathbb R$; the function is not simply arbitrary away from the dyadic rationals. Thus C is false. Therefore, the only statement that must be true is $$ \boxed{\mathrm{D}}. $$